Question: $\log_{9}729 = {?}$
Solution: If $\log_{b}x=y$ , then $b^y=x$ First, try to write $729$ , the number we are taking the logarithm of, as a power of $9$ , the base of the logarithm. $729$ can be expressed as $9\times9\times9$ $729$ can be expressed as $9^3$ $9^3=729$, so $\log_{9}729=3$.